Exhibition of Mathematical Methods English Translation of Sanpo-Hakki 算法発揮 (『大成算経』の数学的・歴史学的研究)

Exhibition of

Mathematical

Methods

English Translation

of

Sanpo-Hakki

算法発揮

Matumoto,

Takao1

(松本堯生) and

Chijiwa,

Tomohiro2

(

千々和智大

)

Professor Emeritus, Hiroshima Univ. Doctor of Science, Hiroshima Univ.

Translators’ Preface

Sanpo-hakki [算法発揮] is the first published book in the world on determinants. It

was

publishedin

1690

atOsaka. Three volumes

are

devotedto thetheoryof elimination

using determinants, appliedto the problems with

answers

and formation ofequations.

We

can

see

theoriginaltextsatWasan$DB^{3}$in DegitalCollections ofTohokuUniversity.

This is

an

abridged translationofSanpo-hakki with call number 7.20306.1 ofKano$(\ovalbox{\tt\small REJECT}$

野$)$ collection. Therewe can see one more published Sanpo-hakki with call number 63

of Okamoto-kan (岡本刊) collection which has

an

advertisement of

some

medicine at

the end of volume 2. Japan Academy [日本学士院] has

an

original Sanpo hakki owned

by Endo with claim number 1705 and

one more

but without volume 3. Mathematics

Department of Kyoto University also has a beautiful

Sanpo-hakki4.

In

1710

it

was

published again, deleting the

name

of the author at the first page of each volume.

Wasan Institute [和算研究所] and Koju

Bunko5

[高樹文庫] have this version. So,

we know now at least six complete original texts with two versions. There

are

also

some

written copies and in 1935 Sugaku-koten-shoin [数学古典書院] published copies

producedon a mimeograph.

Thename ofthe author ofSanpo-hakki isexplicitlywritten at each volume

as

IZEKI

Tomotoki [井関知辰] but by the custom ofthat time his teacher SHIMADA Naomasa

[嶋田尚政]

can

be considered to be the actual author.

Thevolume 2 andthe mainpart of volume

3

are

writtenonly byChinese characters,

but the other parts including the volume 1 which may be considered

as

explanations

are written by Chinese characters and Japanese alphabets katakana mixed.

lAdditionalaffiliation: Vice-Directorof Seki-Kowa Mathematical Instituteat YokkaichiUniversity

andLong-termResearcheratRIMSKyotoUniversity, -mail: [email protected]

2Additional affiliation: Yamaguchi-Konanjuniorhigh school,

e.mail: [email protected]

3http:$//dbr.$library.tohoku.$ac.jp/infolib/metapub/G0000002$wasan (Choose Englishifnecessary.)

4http:$//edb$.math.kyoto-u.ac.jp/wasan/109

In Sanpo-hakki the new unknown is mintorduced by the

name

certain unknown [何某

bo] in the volume 1, celestial element [天元 Tengen] in the volume 2 and Tengen with

some

adjective in _{the volume 3. The side-notation [傍書法 bosho-ho] used in}

Sanpo-hakki is different from that ofSeki school. One verticalsegment _indicates the term of

each degree of the unknown _{in vertical order and the side-notation contains not only}

literal coefficients _{but also numerical coefficients exPressed} by _{Chinese characters and}

$\pm$ signs. But in this translation

we use

a horizontal segment _{instead of vertical}

_one

and _{side-notation becomes} up-notation. _{We add sometimes thecorrespondingmodem}

notation

or

phrases in $[]$ like _{$[a+bx+cx^{2}=0]$}

or

[eliminated expression]. Note that

the

same

notation

was

used to represent the formula $a+bx+cx^{2}$ _{and the equation}

$a+bx+cx^{2}=0$. _{Note also that an} equal$sign=$ _andaparenthesis $()$ were not invented

at that time in Japan. The _footnotes

_are

the translators’ explanations _or comments,

while the explanatory notes at thetop of the volume 3 is the author’s.

For

us

the volume 1 is rather easy to

understand

but the

volumes

2 and

3

seem

unfamihar.

So,

we

add

some

comment

as an

appendix _{at the end.}

$0$

Sanpo-Hakki, Foreword

What

occurs

from heaven odd and earth

even

is the number. The world is not

mixed up due to it. Today and the past are also calm. The operation of the nation

is managed _{well and the difference of} people goes without a deadlock. Though the

purification _{keeps changing day and night, the transition oftime does not differ at all.}

It

runs

well according _{to the objects. There is a Person named IZEKI Tomotoki. He}

leamed

Mathematics

from

SHIMADA

Naomasa. He has beenintelligent since he

was

young

and it is hardto explain that in words. He understands theory in

one

step and

develops his wisdom in a half. _{At last, he discovered what the ancients could not find}

and mentionedwhatthepredecessors_{could not say. That is}

_as

_{if Yellow Emperormeets}

Reishu’s reputation. _How

_can

_{Mathematics in Wei-Tang dynasty reach his readiness?}

Now, he edited Sanpo-Hakki in 3 volumes for

_PeoPle.

_{The theory and} application”

are

simply outlined and economically _{detailed. ThePeople around the world, who try}

to opentheir eyes to Mathematics,

can

escape from thepitch-dark

_cave.

It

can

besaid

that he is a person who knows the unearth

reason

and the obvious phenomena. _He,

however, becamethirty years old. Why does he stop here?

Written6

byhokusui lordless samurai Ichijiken

Ichu7

in May

1690

with two seals (一時軒岡西惟中).

6This hne and the next lineare deletedin the 2nd version.

$7$

OKANISmIchu (1639-1711). This forewordwrittenbyanoldhaikupoet [俳人 haijin] is difficult to understand and translate.

1

Sanpo Hakki,

Volume 1

edited by IZEKI Jubeejo Tomotoki

a student of SHIMADA Naomasa

When

we

want to solve the problem, sometimes it is difficult to obtain

an answer

equation [in

an

asking unknown quantity $x$] directly. In such

a

case

considering

every

quantity in thepossible

answer

equation

as

known,

we

introduce

a

new

unknown

quan-tity $[y]$ bythecelestialelement

method8

and form two equations [in $y$] specifying their

coefficients by the

names

of quantities, numerical factor and plus or minus at the side

ofeachdegree of unknown

as

usual9.

Distinguish them

as

theformer equation and the

latter equation. Using them

we

would get

an

answer equationl

[in $x$]. If it is hard to

do it in one step, we repeat it several times and get the

answer

process [to find $x$].

1.1

Case of two quadratic equations

[Let twoquadratic equations in $y$ be given:]11

The former equation $\underline{+a}\underline{+b}\underline{+c}$ $[a+by+cy^{2}=0]$ in the

new

unknown $[y],$

The latter equation $\underline{+d}\underline{+e}\underline{+f}$ $[d+ey+fy^{2}=0]$ in the new unknown $[y].$

Multiplying the latter equation by $a$ in the former equation,

we

obtain $+ad+ae$

$+af$ from which we subtract the equation

$+ad+bd+cd$

, the former equation

multiplied by$d$in the latter equation. Let the

remainderl2

be the

first

equation.

Multiplyingthe former equation by $f$ in thelatter equation, we obtain

$+af+bf$

$+cf$ from which

we

subtract the equation

$+cd+oe+cf$

, the latter equation

multiplied by $c$ in the former equation. Let the

remainder

be the

second

equation.

8The celestial element method is a method to form an algebraic equation inone variable $x$ with

numericalcoefficients. Thismethod enablesusto manipulate polynomialsin$x$; apolynomial$a+bx+$

$cx^{2}+dx^{3}=0$ _is represented by the vector of its coefficients: $\underline{+a}\underline{+b}\underline{+c}\underline{+d}$. In the same way the

unknownquantity$y$is representedbythevector$\underline{+1}$forexampleinthe volume 3.

9This partseems tocomeffom$Hatusbi-sanp\overline{o}$-endangenkai[発微算法演段諺解]. Tanaka’s

S\={o}shiki-ikkan-no-jutsu[双式一貫之術] in Sangaku-funkai [算学紛解] 第1巻has al-most thesamesentence.

1 _{How to getthisequationin}$x$ is the theme of the volume 1.

llThe formerequation $a+by+cy^{2}=0$ isrepresented by a vector $\underline{+a}\underline{+b}\underline{+c}$, where $a,$ $b$, and $c$

arepolynomials in$x$ with numericalcoefficients; the latterequationis represented similarly. In the

originaltext,the authoruseskana(Japanese alphabets) i(イ), ro(D), $ha(\prime\backslash ),$$\cdots$,whicharerendered

by alphabets$a,$ $b,$ $c,$ $\cdots.$

[Quadratic

convertedl3

expressionl4

;]

The first equation in the new unknown $[y].$

$[(ae-bd)+(af-cd)y=0$

Quadratic expressed as $g+hy=0.$]

$Y\overline{o}-\dot{n}tsu$

The secondequationin the

new

unknown$[y].$

$[(af-cd)+(bf-ce)y=0$

expressed

as

$i+jy=0.$]

[Quadratic

eliminatedl5

expression:]

Quadratic $\oplus gj$ [The unknown $y$ is

now

elimited:

In-ritsu $\ominus hi$

$gj-hi=0$

is anequation in the unknown _$x.$]

The solutionis the following: \copyright in the first equationis the

unknownl6

$[y]$ multiplied

by \copyright. _{So, \copyright multiplied by \copyright is the unknown} $[y]$ multiplied by \copyright and \copyright. Move it to

the left-hand

sidel7

as

_Lefl

$[gj=-hjyarrow Lefl]$ _{Hence it is positive} $[+gj].$

$O\iota$ in the second equation is the unknown _$[y]$ multiplied by \copyright. So,

$\iota$ multiplied by

\copyright is the unknown $[y]$ multiplied by \copyright and \copyright. It cancels out the formula

Lefl

at the

left-hand side

_{$[gj-hi=Lefl+hjy=-hjy+hjy=0]$}

_{Hence it is}

_negativel8

$[-hi].$

1.2

Case of two cubic

equations

[Let two cubic equations in $y$ be given:]

The former equation $\underline{+a}\underline{+b}\underline{+c}\underline{+d}$ in the

new

unknown $[y].$

The latter equation $\underline{+e}\underline{+f}\underline{+g}\underline{+h}$ in the new unknown $[y].$

Multiplying the latter equation by $a$ in the former equation, we obtain the equation

$\underline{+ae}+af+ag+ah$ fromwhich we_{subtractthe equation}

_{$+ae+be+ce+de,$}

the former equation multiplied by $e$ in the latter equation. Let $the\overline{rem}a\overline{inder}\overline{be}$the

first

equation.

$13Sanp\overline{(}\succ$hakkiusestheterm $Y\overline{o}$-ritsu[

陽率] but the term Kanshiki[換式],whichmeanstheconverted

equations, used in$Kaifi_{1}kudai$-n$\triangleright$ho-[解伏題之法] seemsbetter forus tounderstand.

14The encircled one in the table, for example \copyright indicates that $g$ is defined by the above, i.e.,

$g:=ae-bd.$

15Sanpo-hakki uses the term In-ritsu $[$陰率$]$. We find that the In-ritsu equation is obtained by

$e\lim\dot{n}$atingtheextra unknown

$y$fromtwoequationsin$y$

.

In fact, thisisthedeterminant of the above

$Y\overline{o}\dot{n}tsu$ [陽率] matrix and the resultant of the

formerand latterequations.

16Thesigns areignored. _{In the modem notation $g=-hy.$}

17Theoperation isoriginaJlydoneon thecountingboard. When the formulaonthe board is saved

inamemorycalled theleft-hand side, the boardisreset tostart a new oPeration.

l8Ifa fomula $A$ is saved in the left-hand side and a formula _{$B$ is} on _the

board, anew equation $A-B=0$ canbeformedby cancellation.

Multiplying the latter equation by $b$ inthe former equation,

we

obtain the equation $+be+bf\underline{+bg}\underline{+bh}$fromwhich

we

subtract the equation $+af\underline{+bf}\underline{+cf}\underline{+df},$

the former equation multiplied by$f$ in the latter equation. Then, addthefirst equation

to the $remainder_{+be}^{-af}$ $-cf-df+bg+bh$ . Let the sum be the second equation.

Multiplying the $form\overline{ereq}u\overline{ation}\overline{byh}$in the latter equation,

we

obtain the equation

$+ah+bh+ch+dh$

from which

we

subtract the equation

$+de+df+dg$

$\overline{+dh}$, the latter equation multiplied by $d$in the former equation. Let the remainder

be the third equation.

[Cubic converted expression:]

The first equation

in the

new

unknown

$[y].$

The secondequation

Cubic

$Y\overline{o}$-ritsu in the

new

unknown $[y].$

The third equation

in the new unknown $[y].$

[Cubic eliminated expression:]

$\oplus 3$ terms: $[imq+jno+klp$

Cubic

In-rtsu $\ominus 3$ terms:

$-inp-jlq-kmo=0$

]

The solution is the following: To the

coefficientsl9

of the first equation

we

attach

their signsas$\underline{+i}-j+k$. Then, we canget the cubic In-ritsu by multiplying them

by quadratic In-ritsu.

In the following three figures the letters in letters out of

be understood by comparingthem.

Reciting the quadratic

In-ritsuaccordingto the _$+mq$

letters out of

write the right figure by $-np$

using the lettersin $.$

$\dot{\fcircle}\sim+$ $\mu$コ $+imq$ $\frac{\tilde{\frac{i}{\vee\underline{a}}}}{\Xi^{i3}}$ $-inp$ $\ominus\prime.$ $+lq r\circ^{\backslash }\triangleright -jlq$

Thesame asabove.

-no

$\frac{h}{.\underline{a}}$ $+jno$

冒

$+lp \hat{o} +klp$

Thesame as above.

$-mo \frac{\triangleright}{s^{\ni}\primerightarrow\underline{\approx}}\backslash -kmo$

The sum of three terms with $+sign$ and the sum of three terms with $-sign$ are

equal. Thus, this is the cubic In-ritsu.

Although it is clear by the above figures, those who doubt it should understand it

as

follows:

Multiplying the second equation by $\iota$ in the first equation,

we

obtain the

equa-tion $\underline{+il}+im\underline{+in}$ from which we subtract the equation $\underline{+il}\underline{+jl}+kl$ , the

first equation multiplied by の in the second equation. Let the remainder be the top

equation.

Multiplying the third equation by $\iota$ in the first equation,

we

obtain the equation

$\underline{+io}\underline{+ip}\underline{+iq}$ from which we subtract the equation

$+io+jo+ko$

, the first

equation multiplied by $0$ in the third equation. Let the remainder be the bottom equation.

The top equation _{in the new unknown} $[y].$

The bottom equation in the

new

unknown $[y].$

We obtain the following equation from quadratic In-retsu. The multiplication of

$+$ jklo

$-$ ijlq

$g$ and $j$ together is

$-$ ikmo Move the sum to the left-hand side as

Lefl.

The

$+$ jklo

multiplicationof$h$and$i$ together is iklp The

sum

cancelsout the formula

Lefl

ijno

$+$ iinp

$- O\iota jlq$

$\iota kmo$

at the left-hand side. Hence

we

subtract it from

_Lefl

and get $+$ $\iota imq$

We obtain

$+ O\iota klp$

$+$

_{$- O\iota inp$}

the cubic In-ritsu by dividing all of them by $O\iota.$

Althoughwe know how to obtain the In-ritsu in thisway, it becomes complicated in

the

case

of higher degree. We, therefore,calculate the In-ritsu by the previousmethod.

Since

we can

obtain the In-ritsu of arbitrary degree in the

same

way,

we

omit this kind

of explanation in the

cases

of 4th, 5th and 6th degree. We follow only the previous

method.

1.3

Case of two quartic

$2$

equations

[Let two quartic equations in $y$ be given:]

The former equation $\underline{+a}\underline{+b}\underline{+c}\underline{+d}\underline{+e}$ in the

new

unknown $[y].$

The latter equation $\underline{+f}\underline{+g}\underline{+h}\underline{+i}\underline{+j}$ in the

new

unknown $[y].$

From the equation obtained by multiplying the latter equation by $a$ in the former

equation we subtract the equation obtained by multiplying the former equation by $f$

inthe latter equation. Let the remainder be the

_first

equation.

$\mathbb{R}om$ the equation obtained by multiplying the latter equation by $b$ in the former

equation we subtract the equation obtained by multiplying the former equation by $g$

inthe latter equation. Then, add the first equation to the remainder. Let the

sum

be

the second equation.

From the equation obtained by multiplying the latter equation by $c$ in the former

equation we subtract the equation obtained by multiplying the former equation by $h$

in the latter equation. Then, add the second equation to the remainder. Let the

sum

be the third equation.

From the equation obtained by multiplying the former equation by $j$ in the latter

equation

we

subtract the equationobtained by multiplying the latter equation by $e$ in

$2$

the former equation. Let the remainder be the

_fourth

equation.

[Quartic convertedexpression:]

The first equation

in the

new

unknown $[y].$

The second equation

in the new unknown $[y].$

Quartic $Y\overline{o}$-nttsu

Thethird equation

in the new unknown $[y].$

Thefourth equation

in the

new

unknown $[y].$

[Quartic eliminated expression:]

$\oplus 12$ terms:

Quartic

$In-\dot{n}tsu$

$\ominus 12$ terms:

Thesolution is the following: Tothe coefficientsof the first equationwe attach their signs

as

$+$ \copyright – の

$+@-$

. Then, we

can

get thequarticIn-ritsu by multiplying them by cubic In-ritsu.

In the following four figures the letters in $\circ$ are those in the quartic $Y\overline{o}$-ritsu. The

letters out of

$+puz +k\psi uz$

Reciting the cubic In- $+qvx$ $\otimes$

$+kqvx$

ritsu accordingto the let- _$+rty$ $+\triangleright,$

$+krty$

$0$

ters out of

the right figure by using $-pvy$ $\overline{\underline{\circ}}.$ $-k$$pvy$

the letters in$.$ $-qtz$ $s^{\Leftrightarrow}$ -kqtz

-krux

$-rux$

$+ouz$ -louz

$\dot{\ominus}$

$+qvw$ $|$ -lqvw

$+rsy$ $0\succ$

.

-lrsy

The same asabove.

$-\sigma uy \tilde{\frac{}{\underline{a}}} +lovy$ $-q\mathcal{S}Z \dot{5} +lqsz$

$-ruw$ @ $+lruw$

$+otz +motz$

$+pvw \bigotimes_{+} +mpvw$

$+rsx 0\triangleright, +mrsx$

Thesame asabove.

$-rtw-psz-ovx @ \frac{\triangleright}{\dot{H}\underline{a}}\backslash ---mrtwmpszmovx$

$+oty$ $-$ noty

\copyright

$+I^{yuw}$

$|$ -npuw

$+qsx$ $0*$ -nqsx

The same as above. _-our $\underline{\triangleright}_{\backslash } +noux$

$\underline{a}$

$-psy 3rightarrow +npsy$

$-qtw$ @ $+nqtw$

Thesum of12 terms with $+sign$and the

sum of 12 terms with $-sign$ are equal.

Thus, this is the quartic In-ritsu.

1.4

Case of two quintic equations

[Let two quintic equations in$y$ be given:]

The former equation $\underline{+a}\underline{+b}\underline{+c}\underline{+d}\underline{+e}\underline{+f}$ in the new unknown $[y].$

The latter equation $\underline{+g}\underline{+h}\underline{+i}\underline{+j}\underline{+k}\underline{+l}$ in the

new

unknown $[y].$

$\mathbb{R}om$ the equation obtained by multiplying the latter equation by $a$ in the former

equation, we subtract the equation obtained by multiplying the former equation by $g$

in the latter equation. Let the remainder be the

_first

equation.

From the equation obtained by multiplying the latter equation by $b$ in the former

equation, we subtract the equation obtained by multiplying the former equation by $h$

in the latter equation. Then, add the first equation to the remainder. Let the sum be

$\mathbb{R}om$ the equation obtained by multiplying

the latter equation by $c$ in the former

equation we subtract the equation obtained by multiplying the former equation by $i$

in the latter equation. Then, add the second equation to the remainder. Let the

sum

be the third equation.

$\mathbb{R}om$ the equation obtained

by multiplying the latter equation by $d$ in the former

equation we subtract the equation obtained by multiplying the former equation by$j$

in the latter equation. Then, add the third equation to the remainder. Let the

sum

be the

_fourth

equation.

$\mathbb{R}om$ the equation obtained by multiplying the

former equation by $l$ in the latter

equation we subtract the equation obtained by multiplying the latter equation by $f$ in

the former equation. _{Let the remainder be the}

_fiflh

_equation.

[Quintic converted expression:]

The first equation

inthe newunknown $[y].$

The secondequation

in the newunknown $[y].$

Quintic The thirdequation

$Y\overline{o}$-ritsu

in the newunknown $[y].$

The fourth equation in the new unknown $[y].$

The fifth equation in thenewunknown $[y].$

Quintic In-ritsu $\oplus 60$ terms:

msyEK mszFI msADJ $mtxFJ$

mtzCK mtAEH muxDK $muyFH$

muACI mvxEI mvyCJ $mvzDH$

nryFJ nrzDK

nrAEI

$ntwEK$

ntzFG

ntABJ

nuwFI $nuyBK$

nuADG nvwDJ nvyEG $nvzBI$

orxEK orzFH $\sigma rACJ$ $oswFJ$

oszBK osAEG

cxuw

$CK$ $ouxFG$

ouABH

ovwEH $\sigma vxBJ$ $ovzCG$

prxFI pry$CK$ prADH $pswDK$

psyFG psABI ptwFH $ptxBK$

ptACG pvwCI pvxDG $pvyBH$

qrxDJ qryBH qrzCI $qswEI$

qsyBJ qszDG qtwCJ $qtxEG$

qtzBH quwDH quxBI $quyCG$

msyFJ mszDK msAEI $mtxEK$

mtzFH mtACJ muxFI $muyCK$

muADH mvxDJ mvyEH $mvzCI$

nryEK nrzFI nrADJ $ntwFJ$

ntzBK ntAEG nuwDK $nuyFG$

nuABI nvwEI nvyBJ $nvzDG$

orxFJ orzCK orAEH $oswEK$

Quintic In-rtsu $\ominus 60$ terms: oszFG osABJ ouwFH $(yuxBK$

ou

$ACG$ ovwCJ

ovxEG

$ovzBH$

prxDK pry$FH$ prACI $pswFI$

psyBK ps$ADG$ ptwCK $ptxFG$

ptABH $I^{yvwDH}$ pvxBI $pvyCG$

qrxEI qryCJ qrzDH $qswDJ$

qsyEG qszBI qtwEH $qtxBJ$

qtzCG quwCI quxDG $quyBH$

(We omit to tmnslate the solution with 5 figures similar to the case

_of

quartic

equa-tions except the following last sentence.)

The

sum

of

60

terms with $+sign$ and the

sum

of 60 terms with $-sign$ are equal.

1.5

Case

of

two

sextic

equations

[Let two sextic equations in $y$ be given:]

The former equation $\underline{+a}\underline{+b}\underline{+c}\underline{+d}\underline{+e}\underline{+f}\underline{+g}$ in thenew unknown $[y].$

The latter equation $\underline{+h}\underline{+i}\underline{+j}\underline{+k}\underline{+l}\underline{+m}\underline{+n}$ in the new unknown _$[y].$

Since howto obtain the $Yo$-ritsu is the same

as

before,

we

omit it.

The first equation. The second equation. The third equation. The fourth equation. The fifth equation. The sixth equation.

$9_{n-\gamma\eta tsu}^{ext_{!}c}$

By the diagram of sextic $Yo$-Witsu

above

we

make six figures and then

$\oplus 360$ terms. obtain the sextic In-ritsu by

multi-plying by the quintic $In-\dot{m}tsu$. The

$\ominus 360$ terms _{methodis the}

same as

inthe

case

of

quadratic, cubic, quartic

or

quintic

In-ritsu. Thus,

we

omit the details.

$\bullet$ We omit the

case

when the formerand latter equations

are

ofdegree greaterthan

6. You

can

guess the method through the above examples. Although there

are

other methods to obtain the In-ritsu directly besides two methods given in this

book,

we

do not mention about them because it is not easy for the beginners

to understand. After understanding this book,

you

can

find out the methods by

yourselves.

$\bullet$ As stated at first, after finding the former and latter equations you will get the

answer

equation that

means

the equation in the true unknown by using $Y\overline{o}-$

rtsu and In-ritsu [the converted andeliminated expressions] for the appropriate

degree. Now for the better understandingofthe beginners

we

add the problems

and their

answers

in the next volume and how to form the equations in the

final volume. The able people should not follow the details and try to solve by

themselves. Even with three volumes, which teach rules, answer processes and

how to form the equations, it might be difficult for the usual people to

understand

without oral teaching. But if you

never

give up by the hardness and read them

every day and night, youwillunderstand them someday.

2

$Sanpo-$

Hakki,

Volume

₂

_{edited by IZEKI Jubeejo Tomotoki}

a student of SHIMADA Naomasa

2.1

Problem

1

[Problem:] There is a rectangle inside a

right-angled triangle as shown in the figure. Let $A$ be

the

sum

of the hypotenuse [Gen, 弦] and the

area

$[Gai,$ _外$]$ outside the

rectangle.21

Let _$B$ be the

sum

of the height $[Koh,$ 勾$]$ and the long side $[Cho,$ 長$].$

Let $C$be the

sum

of the base _{$[Ko,$ 股}$]$ and the short

side $[Hei,$ _平$]$. Find the values of_{$Koh,$ $Ko$}, Gen, _$Cho$

and $Hei.$

Answer: By the process stated below, we can find $Koh.$

Answerprocess; By the method of celestial element, let $Koh$ be the unknown.

Sub-tracting it from $B$, we get $Cho$ as the remainder $[Cho=B-Koh]$_{. Subtract it from}

$C$ and multiply the remainder by _$Koh$

.

_{Call the product}

$\alpha[\alpha=Koh\cross(C-Cho)]^{22}$

Take $Koh$ and add $C$ to it. Call the

sum

_{$\beta[\beta=Koh+C]^{23}$}

Take 2 times$A$, from whichwesubtract _$C$ multiplied by _$Koh$

and call the remainder

$\gamma[\gamma=2A-Koh\cross C].$

Take $Koh$ and add 2 times $Cho$ to it. Call the

sum

_{$\delta[\delta=Koh+2Cho].$}

Take the

sum

of 4 times $Koh$ squared24 _{and 4 times} $C$ squared, from which we

subtract $\gamma$ squared and call the remainder _{$\epsilon[\epsilon=4Koh^{2}+4C^{2}-\gamma^{2}].$}

Take the

sum

of8 times $C$ and 2 times $\delta$ multiplied

$\gamma^{25}$ Call it _{$\zeta[\zeta=8C+2\delta\cross\gamma].$}

Take$\delta$ squared,

from which we subtract 4

rods26

and call theremainder$\eta[\eta=\delta^{2}-4].$

2lIntheoriginaltext, givenquantities_{arecalled tada-iu-su(只云数), mata-iu-su(又云数) and}

betsu-$iu$-su-(別云数), whichare rendered_{by$A,$ $B$ and$C.$}

22Intheoriginaltext, the authorusesthenameof28 constellationskaku$(\S),$$k\overline{o}(\hat{f|_{d}}),$ $tei(fi),$ $b\overline{o}(\overline{E})$, shin$(J\backslash \llcorner\backslash ),$ _{$bi(\not\in),$}$ki(g),$

$kei(=\Rightarrow f)$ (Theauthorusesthis instead of

$tou(\backslash j^{\backslash }).$), $gy\vec{u}(*),\check{J}^{o(x)},$$\cdots$, which

arerenderedbygreek _alPhabets $\alpha,$$\beta,$ $\gamma,$ $\delta,$

$\epsilon,$ $\zeta,$ $\eta,$ $\theta,$

$\iota,$ $\kappa,$ $\cdots.$

23The first operation _is originally _{to arrange a} Polynomial on the countingboard. But there is

no comment on the arrangement in the caseoftakin$g$the 4 sum. So, to takeisused for every first

operation inthis translation.

24We use the term $A$ squared rather than the square of _$A$,

because it is closer to the original

expressionA巾$.$

25Laterwewillusethe expression 2 times theproductof$\gamma$ and

$\delta$’ aswell.

Take the product of $\beta$ and $\epsilon$, from which

we

subtract the product of

$\alpha$ and

.

Call

the

remainder

$\theta[\theta=\beta\cross\epsilon-\alpha\cross\zeta].$

Take the product of $\alpha$ and $\eta$ and add

$\epsilon$ to it. Call the sum $\iota[\iota=\alpha\cross\eta+\epsilon].$

Take the product of $\beta$ and $\eta$ and add

$\zeta$ to it. Call the

sum

$\kappa[\kappa=\beta\cross\eta+\zeta].$

Take the product of$\kappa$and $\theta$ and

move

it to the

left-hand

side

as

Lefl

$[Lefl=\kappa\cross\theta].$

Take $\iota$ squared. Since it cancels out [the formula]

Lefl

at the left-hand side,

we

obtain an equation $[Lefl-\iota^{2}=0]$

.

Solving the equation of degree 8, we obtain $Koh.$

It

answers

the problem.

2.2

Problems 2

to

6

(We omit to translate the subsections 2.2-2.6.)

2.7

Problem

7

[Problem:] There is a pentagon inside a circle as shown

in the figure. When the length offive sides$a,$ $b,$ $c,$ $d$ and $e$

are given, find the diameter ofthe circle.

Answer: By the process stated below, we can find the

diameter of the circle.

Answer

process:

By the celestialelement method let the

diameter ofthe circle be the unknown. Its square is called

$\alpha.$

Take the 4thpower of$b$ and add the 4thpower of $c$ to it. $\mathbb{R}om$ the

sum

subtract 2

times the product of $b$ squared and $c$ squared. Multiply the remainder by $\alpha$, and call

the product $\beta.$

Take the product of$\alpha$ and $b$ squared and add the product of

$\alpha$ and $c$ squared to it.

From the sum we subtract 2 times the product of $b$ squared and $c$ squared, and call

the remainder $\gamma.$

Take the 4thpowerof$d$andaddthe 4thpower of$e$ to it. $\mathbb{R}om$the

sum

we

subtract

2 times the product of$d$squared and $e$ squared, and call theremainder

$\delta.$

Take the product of$\alpha$ and $d$ squared and add the product of

$\alpha$ and $e$ squared to it.

From the

sum we

subtract 2 times the product of $d$ squared and $e$ squared, and call

the remainder $\epsilon.$

Take $\alpha$ from which we subtract 2 times $a$squared, and call the remainder

Take2 times theproduct of$\alpha$squared and _$a$ squared and add2 times the productof

$\epsilon$ and $\zeta$toit. $\mathbb{R}om$ thesum wesubtract theproductof

$\alpha$and $\gamma$, and call the remainder

$\eta.$

Take the 4 sum of the 6 times the product of$\alpha$ squared and the 4th power of

$a,$ $8$

times the product of $\epsilon,$ $\zeta$ and $a$ squared, 4 times the product of $\delta$ and

$\zeta$ squared and

4 times $\epsilon$ squared. From the

sum

we subtract the

sum

of 2 times the product of $\alpha$

squared and $\delta,$ $4$ times the productof

$\alpha,$ $\epsilon$ and _$a$squared and the product of

$\alpha$ and $\beta.$

Then, we call the remainder $\theta.$

Take the 4 sum of the product of $\alpha$ squared and the 6th power of _$a$, the product

of $\delta,$ $\epsilon$ and $\zeta$, the product of

$\epsilon,$ $\zeta$ and the 4th power of _$a$ and 2 times the product

of$\epsilon$ squared and

$a$ squared. $\mathbb{R}om$ the sum we subtract the sum of the product of

$\alpha$

squared, $\delta$ and

$a$ squared, 2 times the product of $\alpha,$ $\delta,$ $\zeta$ and $a$ squared and 2 times

the product of$\alpha,$ $\epsilon$ and the 4th power of

$a$. Then, we call the remainder $\iota.$

Take the 4

sum

of theproductof$\alpha$ squared and $\delta$ squared, the productof

$\alpha$squared

and the8th power of$a,$ $4$times the product of$\epsilon$ squared and the 4thpower of

$a$ and 2

times the product of $\alpha$ squared, $\delta$ and the 4th power of

$a.$ $\mathbb{R}om$ the

sum

we subtract

the

sum

of4 times the product of$\alpha,$ $\delta,$ $\epsilon$ and _$a$squared and 4 times the product of

$\alpha,$

$\epsilon$ and the 6th power of

$a$. Then, we call the remainder $\kappa.$

Take 4 times the product of $\beta$ and $\iota$ from which we subtract 2 times the product of

$\gamma$ and $\kappa$, and callthe remainder $\lambda.$

Take the product of $\beta$ and $\theta$ from which we subtract the

product of $\alpha$ and $\kappa$, and

call the remainder $\mu.$

Take the product of $\beta$ and

$\eta$ and add the product of$\gamma$ and

$\theta$.to it. $\mathbb{R}om$ the sum

we subtract 2 times the product of$\alpha$ and $\iota$, and we call the remainder

$\nu.$

Take the

sum

of the product of$\alpha$ and $\mu$ squared, 4 times the product of$\beta$ squared,

$\eta$ and $\nu$ and 8 times the product of

$\gamma$ squared, $\eta$ and

$\lambda$ and move it

to the left-hand

side

as

_Lefl.

Take the

sum

of 2 times the product of $\alpha,$ $\nu$ and $\lambda$ and 8 times the product of

$\beta,$

$\gamma,$ $\eta$ and $\mu$. Since it cancels out [the formula]

Lefl

at the left-hand side,

we

obtain an

equation. Solving the

equation27

of degree 14,

we

obtain the diameter. It

answers

the

problem.

The end of Sanpo-Hakki Volume 2.

3

Sanpo

Hakki,

Volume

3

3.0

Explanatory

notes

$\bullet$ The changes of plus and minus, addition and subtraction

are as

usual.

$\bullet$ Each encircled

name

oftwenty-eight

constellations28

below29

the

horizontal

seg-ment representstheplacewhere the formula

described

above the

horizontal

segment –

are

moved to. Even if there

are

2, 3, 5

or

10

names

above

–,

we

abbreviate them into

one

name

to simplify the sentences. In many books we find

“寄何位”30 but we omit “位” to simplify the sentences in the volume 2 of this book. $+1KohCho$

$\bullet$ There is $-1KohC$ inthe formerequation ofProblem 1. It is

an

abbreviation

of$Koh$侃$o$ Multi. 1$+$ [勾長相乗一段正] and $KohC$ Multi. 1-[勾別云数相乗一段

負$]$; we omit the letters Multi. [相乗], the unit [段] and the value [云数] in the

value $C$ [別云数]. The letters $Eho$ [法] forEnsekiho[円積法] inProblem 3and $Gai$

[外] for the

area

ofcomplement inProblem

5

are

the

same

kind ofabbreviations.

$\bullet$ If there is a

common name

in all the coefficients of the former or latter

equa-tion, we

can

omit the

name.

If there is a

common

numerical divisor in all the

coefficients of the former or latter equation, we

can

reduce the

common

divisor.

We

can

applythe

same

to the [converted] $Y\overline{o}-$

or

[eliminated] In-diagrams. This

method isca垣edreduction method of all

common names

and

common

divisors [遍

省遍約之法 Hensh$\overline{(}\succ henyaku-n\not\in$ho]. You willsee this operationin the volume 3.

$\bullet$ In the

answer

process

to Problem 3, we wrote: “Take 8times the product of $Koh$

and $Eho$ from which we subtract $\alpha$, and call the remainder $\beta$

.

“ On the other

$-2$

$+8KohEho$

hand, wehave inthe former equation. Since the sum ofone $Koh$ and

$-1Koh$

$+$ \copyright

number 2 equals $\alpha$, we

can

call $\beta$

as

in the volume 2. This is another method to

simplify the sentences. There is a further example for the term $\theta$ of Problem 3,

and so on.

$\overline{28They}$

arerendered by Greek letters.

29At the left-handsideofthevertical segmentin the original text.

$\bullet$ $In^{31}$Hatsubi-genkai,

Meigen, Ikkyoku-sanpo and others, it iswritten the

answer

of

second toseventh _{power formula but not written how to get it. So, the beginners}

cannot get the higher power formula. Here,

we

show how to get it.

$\triangle$ The second power

formula:

By the celestial element method we take a new unknown $\circ$ $[y]$ and

move

its square

The square of the

new

$unknown^{32}$ is _{supposed to be known and cancels out}

[the formula]

_Lefl

at the left-handside. Hence [we get]

the former equation $\frac{+1fT\mathfrak{o}^{2}}{\fcircle a}$

By the given problem _we get

the

latter

equation $\frac{\pm 実}{@}$ $\frac{\pm法}{Oe}$

in the

same

unknown $[y]$. Thepart of the degree2 and _{greater is amultiple of}

the square of the unknown $[y]$. So, substitutingthe squarewe add it_{by sliding}2

places to$1eft^{33}$ usingplus or minusas _usual.

Applying Caseoftwoquadratic equationsin the volume 1,

we

getthe second

power _formula.

$\triangle$ The third power formula:

By the _{celestial element method we take a new unknown} $O$ $[y]$ and

move

its cube

The cube of the unknown $[y]$ is supposed to beknown and cancels out [the

formula]

_Lefl

at the

left-hand

side. Hence [we get]

the

former

equation $\frac{+1no^{3}}{\fcircle a}$

By thegiven problem _we get

the latter equation $\frac{\pm\not\equiv}{\fcircle e}$ $\frac{\pm\ \backslash }{ }$ $\frac{\pm\ovalbox{\tt\small REJECT}}{@}$

in the

same

unknown $[y]$. The part of the degree 3 _{and greater is a multiple}

of the cube of the unknown $[y]$. So, substituting thecube we add it _{by sliding3}

placesto left, using plus or minus as usual.

Applying Case of two cubic equations _{in the volume 1, we} get the third

power

formula.

$\overline{3lThese}$

namescertainly stand for$Hatsubi-sanp\overline{o}$-endangenkai[発微算法演段諺解] _(1685),

Meigen-anpo [明元算法] (1689) and [一極算法] (1689).

32Thesquare$f\sigma J^{2}$ of the

unknown $y$is usually given bya formulainthetrue unknown

$x.$ 33Ofcourse “up” _{in the}originaltext.

In the

case

ofdegreegreaterthan 3, the method to get

power

formulaisthe

same as

the above examples and we omit further details.

The end of Sanpo-Hakki

Volume34

3:

Explanatory notes.

3Sanp\={o}-Hakki, Volume 3

edited by IZEKI Jubeejo Tomotoki a student of SHIMADA Naomasa

3.1

Problem

1,

Formation

of equations

$\bullet$ The sumof $Gai$ and Genis given

as

$A[A=Gai+$

Gen].

$\bullet$ The

sum

of $Ko$ and $Hei$ is given

as

$C[C=Ko+$

$Hei].$

$\bullet$ $Koh$ is [supposed to be] given. It is the unknown

in the

answer

process $[x=Koh].$

$\bullet$ The $Cho$ is given

as

the remainder

ofthe subtractionof$Koh$ from $B[Cho=B-Koh].$

Temporarily let $Hei$be a newunknown

Subtract it from$C$. Then, the remainderequals$Ko;+1C$ $\underline{-1}[C-y]$. Add $Koh$

$+1Koh$

to it. Then, the

sum

equals the

sum

of $Koh$ and _{$Ko;+1C$ $\underline{-1}[Koh+C-y].$}

$+1Koh$

Multiply it by $Hei$ and

move

the product to the left-hand side;

$+1C$

$\underline{-1}$

$[Lefl=(Koh+C)y-y^{2}].$

Take $C$ and subtract $Cho$ from it. Then, the remainderequals the

sum

of the short

$-1Cho$

$Koh$ and the short _$Ko;+1C$ $[C-Cho]$

.

Multiply it by $Koh$. Then, the product $-1KohCho$

equals35

the product of$Hei$ andthe

sum

of$Koh$ and_$Ko;+1KohC$ $[Koh\cross C-Koh\cross$ $Cho]$

.

Since it cancels out [the formula]

Lefl

at the left-hand side,

we

get

34Thetitle of the volume3 appearsagain in theoriginalbook. So, we followit.

35Bythesimilarityof the right-angled triangleswehave

$+1$ KohCho $+1Koh$

the former equation

$\frac{-1KohC}{-\fcircle\alpha}\frac{+1C}{+\copyright}\underline{-1}$ in the unknown

$Hei.$

$[-\alpha+\beta y-y^{2}=0;\alpha=Koh\cross C-Koh\cross Cho, \beta=Koh+C].$

Take$Ko;+1C$ $-1$ $[C-y]$, and multiply it by $Koh$. Then, the product equals

2 times the

area

of the triangle; $+1KohC$ $-1Koh$ $[Koh\cross C-Koh\cross y]$

.

Move it to

the lst place.36

Take _$Hei$ $[y]$, and multiply it by 2 times _$Cho$. Then, theproduct equals 2

times the

area

ofthe rectangle; $\underline{+200}[2Cho\cross y]$. Add 2 times $A$to it; _$+2A$

$+2Cho$ $[2A+2Cho\cross y]$

.

Subtract _the formula _{at the lst place from the}

_sum.

_Then,

$-1KohC +1Koh$

the remainder equals 2 times Gen; $\underline{+2A}+2Cho$ $[\gamma+\delta y,$ _{$\gamma=2A-Koh\cross$}

$+$ $+$

$C,$_{$\delta=2Cho+Koh]$}. Move its square to the 2nd place;

$\underline{+1\gamma^{2}}+2\gamma\delta$ $+1\delta^{2}$ $[\gamma^{2}+2\gamma\delta y+\delta^{2}y^{2}].$

Take $Ko+1C$ $-1$ $[C-y]$ and make its square $+1C^{2}$ _{$-2C$ $+1$} $+1Koh^{2}$

$[C^{2}-2Cy+y^{2}]$. Add $Koh$ squaredto it. Then, the sum _equals Gen_squared;

$+1C^{2}$

$\underline{-2C}+1$ $[Koh^{2}+C^{2}-2Cy+y^{2}]$. Multiplying it by number 4, the product

cancels out the formula at the 2nd place. So, we get

$+1\gamma^{2}$

$-4Koh^{2}+2\gamma\delta +1\delta^{2}$

the latter equation _{in the unknown} $Hei.$

$\frac{-4C^{2}}{-\fcircle\epsilon}\frac{+8C}{+\copyright}\frac{-4}{+\copyright}$

$[-\epsilon+\zeta y+\eta y^{2}=0;\epsilon=\gamma^{2}-4Koh^{2}-4C^{2}, \zeta=2\gamma\delta+8C, \eta=\delta^{2}-4].$

Hereafter we refer to “Case oftwo quadratic equations” in the volume 1.

The former equation in the unkhown $Hei.$

$\underline{-\alpha}\underline{+\beta}\underline{-1}$

a $b$ _$c$

The latter equation in the unknown $Hei.$

$\underline{-\epsilon}\underline{+\zeta}\underline{+\eta}$

$d e f$

Following the letters given under the segments in the quadratic $Y\overline{o}$-ritsu [converted

expression],

we

obtain the [converted] $Y\overline{o}$-diagram as follows:

36In the original text the author uses the name of ten calender signs $k\overline{o}(tF),$ $otsu(z_{\lrcorner}),$ $hei(R)$,

$\{$

$+\beta\epsilon$ $-\epsilon$

$-\alpha\zeta$ $-\alpha\eta$ The lst equation in $Hei.$

$g:+\copyright$

$h:-$

The [converted] $Y\overline{o}$-diagram:

$-\epsilon$ $+\zeta$

$-\alpha\eta$ $+\beta\eta$ The 2nd equation in $Hei.$

$i\overline{:-\fcircle\iota}$ $j\overline{:+\fcircle\kappa}$

Now following the letters given

un

derthe segments in the quadratic In-ritsu

[elimi-nated expression], we obtain the [elimnated] In-diagram as follows:

The [eliminated] In-diagram: $\{\begin{array}{ll}+gj: +\theta\kappa-hi: -\iota^{2}\end{array}$

The terms with$+sign$inthis diagram

are

moved to the

left-hand

side

as

inthe volume

2 and canceled out by the terms

with-sign.37

3.2

Problems 2

to

6,

Formation

of equations

(We omit to tmnslate the subsections 3.2-3.6.)

3.7

Problem 7,

Formation of equations

$\bullet$ _$a$ is

given.38

$\bullet$ $b$is given.

$\bullet$ $c$is given.

$\bullet$ $d$is given. $\bullet$ $e$ is given.

$\bullet$ The diameter of the disk is [supposedto be] given.

It is the unknown $[x]$ in the

answer

process.

$\bullet$ Thesquare $[x^{2}]$ of the diameter of the disk is given.

It will be denoted by $\alpha.$

$\overline{37The}$

resultant is given bythe equation$\theta\kappa-\iota^{2}=0$ in the original unknown$x.$

$0$ Since it is difficult to findthe formulaswhich cancel

out each other at once,

we

assume

that $A$ squared39 is given.

First, let$B$ squaredbea

new

unknown

celestial element method. Add$d$squaredtoit andweget

$\underline{+1d^{2}}+1$ from whichwesubtract $e$squared. Then,

the remainder equals 2 times the product of $B$ and $C$;

$\underline{+1d^{2}-1e^{2}}+1$ Its square $-2d^{2}e^{2}$ $-2e^{2}$

$+1e^{4}$

$\underline{+1d^{4}} +2d^{2} \underline{+1}$

equals 4 times the product of$B$ squared and $C$ squared. Move it to Heaven.

Take 4 times $d$ squared

$\underline{+4d^{2}}$ and multiply it by $B$ squared _$d^{2}4+$ . From $-1e^{4}$

the product we subtract the formula at Heaven and get $+2d^{2}e^{2}$ $+2e^{2}$

$\underline{-1d^{4}} +2d^{2} \underline{-1}$ ’

which equals4 4 times the product of $B$ squared and $D$ squared. Multiply it by the

$-1\alpha e^{4}$

square of the diameter and we get $+2\alpha d^{2}e^{2}$ $+2\alpha e^{2}$

and

move

it to Left.

$-1\alpha d^{4} +2\alpha d^{2} -1\alpha$

Take $d$ squared and multiply it by

$e$ squared $+1d^{2}e^{2}$ Then, the product equals

the product ofthe square ofthe diameter and $D$ squared. Multiply it by 4 times $B$

squared and we get $+4d^{2}e^{2}$ which cancels out _{the formula at Left. [So, we get]} $-1\alpha e^{4} -4d^{2}e^{2}$

$+2\alpha d^{2}e^{2} +2\alpha e^{2}$

the first former equation

$\frac{-1\alpha d^{4}}{+1\alpha\copyright} \frac{+2\alpha d^{2}}{+1\otimes} \underline{-1\alpha}$

in the unknown $B$ squared.

Here \copyright and \copyright

are

the formulae which will be moved to the lst place and the 2nd

place respectively in the second latter process.

$\mathbb{R}om$ the

sum

of_$B$ squared and _$A$

squared $+1A^{2}$ $+1$ we subtract $a$ squared.

Then, the remainder $\underline{+1A^{2}-1a^{2}}+1$ equals 2 times the product of

$B$ and $E$. Its

$+1a^{4}$

square $-2a^{2}A^{2}$ _$-2a^{2}$

equals 4 times the product of $B$ squared and $E$ $\underline{+1A^{4}} \underline{+2A^{2}} \underline{+1}$

$39A,$$B,$$C,$$D,$ $E,$$F,$$G$and $H$as wellas$a,$$b,$$c,$ $d,$$e$areshowninthe figures.

4 _{By the similarity oftwo right-angled}

squared. Move it to Earth.

Take 4 times $A$ squared $+4A^{2}$ and multiply it

by $B$ squared; $\circ$ $+4A^{2}$ From the product we

subtract the formula at Earth and get the

remain-$-1a^{4}$

der $+2a^{2}A^{2}$ $+2a^{2}$ which equals 4 times $-1A^{4}$ $+2A^{2}$ $\underline{-1}$

the product of $B$ squared and $F$ squared.

Multi-ply it by the square of the diameter and

we

get

$-1\alpha a^{4}$

$+2\alpha a^{2}A^{2}$ $+2\alpha a^{2}$ which

we move

to Left 2.

$-1\alpha A^{4}$ $+2\alpha A^{2}$ $-1\alpha$

Take$a$squaredand multiply it by $A$squared. Then, we get $+1a^{2}A^{2}$ , which equals

the product of the square of the diameter and $F$ squared. Multiply it by 4 times $B$

squared and

we

get $+4a^{2}A^{2}$ which cancels out the formulaat Left 2 [and

we

get]

$-1\alpha a^{4} -4a^{2}A^{2}$ $+2\alpha a^{2}A^{2} +2\alpha a^{2}$

the first latter equation $-1\alpha A^{4} +2\alpha A^{2} -1\alpha$

$+1\alpha$ \copyright $+1\otimes$

in the unknown $B$ squared.

Here \copyright and \copyright are the formulas which will be moved to the 3rd place and the 4th

place respectively in the second latter process.

Referringto “Case of two quadratic equations” inthe volume 1 with the first former

and latter equations,

we

obtain the first [converted] $Yo$-diagram and the first

[elimi-nated] In-diagram

as

follows:

一1\copyright \copyright $+1\alpha\copyright$

Making the $Yo$-diagram

$\frac{+1\copyright\copyright}{\copyright}\frac{-1\alpha\copyright}{\copyright}$

as

usual and dividing all

The first $Yo$-diagram [the coefficients] by$\alpha$,

we

$+1\alpha\copyright$ $+1$

obtain this simplified $Y\overline{o}-$

The first In-diagram $+1\otimes^{2}\copyright$ $+1$ $\copyright^{2}$ $+2\alpha^{2}$ ´ $-1$ _↓ $-1$ _´↓ $-1\alpha^{2}\copyright^{2}$ $-1\alpha^{2}\copyright^{2}$

The

sum

of the terms with

$+sign$will be the formula

at the 5th place and the

sum

of the terms with

-$sign$will cancel itout [and

they will form the second

latterequation].

Second, let $A$ squared be another unknown $\circ$ Add $b$ squared to it and

we get the formula $\underline{+1b^{2}}+1$ from which we subtract $c$ squared. Then, the

remainder equals 2 times the product of $A$ and $G$; $+1b^{2}-1c^{2}$

$+1$ Its square

$+1c^{4}$

$-2b^{2}c^{2}$ $-2c^{2}$

$+1b^{4}$ $+2b^{2}$ $\underline{+1}$ equals 4 times the product of

$A$ squared and $G$ squared.

Move it to Top.

Take 4 times $b$ squared $+4b^{2}$ and multiply it by

$A$ squared. Then, we get

we subtract the formula at $T_{oP}^{-}$_{The remainder}

$-1c^{4}$

$\underline{+2b^{2}c^{2}-1b^{4}}$ $+2b^{2}+2c^{2}$

$\underline{-1}$

equals4 times the product of

$A$squared and$H$squared. Multiply it by the square of the _{Figure for the second}

$-1\alpha c^{4}$

former equation

$+2\alpha b^{2}c^{2} +2\alpha c^{2}$

diameter and

we

get _{$-1\alpha b^{4} +2\alpha b^{2} -1\alpha$} wh\’ich

we

move

to Bottom.

Take $b$ squared and multiply it by

$c$ squared. Then, the product equals the product

of the square of the diameter and $H$ squared $+1b^{2}c^{2}$ Multiply it by 4 times $A$

squared and we get $+4b^{2}c^{2}$ It cancels out the formula at Bottom _[and we _get]

一1$\alpha c^{4}$ _{$-4b^{2}c^{2}$}

$+2\alpha b^{2}c^{2} +2\alpha c^{2}$

the second former equation _{$-1\alpha b^{4} +2\alpha b^{2} -1 \alpha$}

$-1\otimes +2\otimes$ Named$LC[\mathbb{R}]$

Hereafter,

we

will

concern

the

process

to get the

second

latter equation. Refer to

the first In-diagram above.

Taking the 3 sum of-l times the 4th power of $d,$ $2$ times the product of$d$ squared

and $e$ squared and $-1$ times the 4th power of $e$, we get the number [the formula]

$-1e^{4}$ $+2d^{2}e^{2}$

$-1d^{4}$ Move it to the lst

place.41

$-1$ \copyright

Takingthe 3

sum

of2 times the productofthe squareofthe diameterand $d$squared,

2 times the product of the square of the diameter and $e$ squared and $-4$ times the

$-4d^{2}e^{2}$

$+2\alpha e^{2}$

product of$d$squared and $e$ squared, we get

$+2\alpha d^{2}$ which we

move

to the 2nd place.

$+2$

Taking the 3

sum

of-l times the 4th power of$A,$ $2$

, times the product of

$a$ squared

and $A$ squaredand-l times the 4thpowerof$a$,

we

get $-1a^{4}$ $\underline{+2a^{2}}\underline{-1}$which

we move

to the 3rdplace.

Taking the

3

sum

of 2 times the product of square of the diameter

and

$A$ squared,

2 times the product of the square of the diameter and $a$ squared and $-4$ times the

$-4a^{2}$

product of$a$squared and $A$ squared, weget

$+2\alpha a^{2}$ $+2\alpha$ which

we move

to the

$-\overline{+2\copyright}$

4th place.

Taking the3

sum

of the product of thesquareof theformulaat the 2nd placeandthe

formula at the 3rdplace $-4\epsilon^{2}a^{4}$ $+S\epsilon^{2}a^{2}$ $-4\epsilon^{2}$ , the product of the fomula at

the lst place and the square of the formula at the 4th place $-4\alpha^{2}\delta a^{4}$ $-8\alpha\delta\zeta a^{2}$

$-4\delta\zeta^{2}$ and 2 times the product ofthe 4th power of the diameter, the formula at

the lst place and the formula at the 3rd place $+2\alpha^{2}\delta a^{4}$ $\underline{-4\alpha^{2}\delta a^{2}}\underline{+2\alpha^{2}\delta}$ ,

$+8\epsilon^{2}a^{2} -4\epsilon^{2}$

we get $-4\epsilon^{2}a^{4}-2\alpha^{2}\delta a^{4}$ $\underline{-8\alpha\delta\zeta a^{2}-4\alpha^{2}\delta a}^{2}+2\alpha^{2}\delta-4\delta\zeta^{2}$ which we

move

to the 5th place.

Takingthe 4sumofthe product of the formulae at the2nd, the3rd and the 4th places

$+8\alpha\epsilon a^{4}-4\alpha\epsilon a^{2}$

$-4\alpha\epsilon a^{6}-4\epsilon\zeta a^{4}+8\epsilon\zeta a^{2}-4\epsilon\zeta$ , the product of theformulasat the lst,the 2nd

and the 4thplaces $-4\alpha\delta\epsilon a^{2}-4\delta\epsilon\zeta$ , the product of the 4th power ofthe diameter

and the squareofthe formula at the 3rd place $+1\alpha^{2}a^{8}$ $-4\alpha^{2}a^{6}$ $+6\alpha^{2}a^{4}$ $-4\alpha^{2}a^{2}$

$+1\alpha^{2}$ and the product of the 4th power of the diameter and thesquare of the

for-4lInthe original text the author usesthe nameof animal zodiac ne$(\mp),$ $u/shi(E),$ $tom(\not\in),$ $u(g\beta)$,

$-4\alpha\epsilon a^{6} +8\alpha\epsilon a^{4}$

$-4\alpha\delta\epsilon a^{2}-4\epsilon\zeta a^{4} -4\alpha\epsilon a^{2}$

mulaat the lst place $+1\alpha^{2}\delta^{2}$ ,

we

get $+1\alpha^{2}a^{8}$ _{$-4\delta\epsilon\zeta$} $+8\epsilon\zeta a^{2}$ $-4\epsilon\zeta$

$+1\alpha^{2}\delta^{2} -4\alpha^{2}a^{6} +6\alpha^{2}a^{4} -4\alpha^{2}a^{2} +1\alpha$

which cancels out the formula at the 5th place.42 [So, we get]

$+8\epsilon^{2}a^{2}$

$-4\epsilon^{2}a^{4} -S\alpha\delta\zeta a^{2}-4\epsilon^{2}$

$-2\alpha^{2}\delta a^{4} -4\alpha^{2}\delta a^{2} -4\delta\zeta^{2}$

the second $+4\alpha\epsilon a^{6}$ $-8\alpha\epsilon a^{4}$ $+2\alpha^{2}\delta$

latter equation $+4\alpha\delta\epsilon a^{2}$ _{$+4\epsilon\zeta a^{4}$} $+4\alpha\epsilon a^{2}$

$-1\alpha^{2}a^{8} +4\delta\epsilon\zeta -S\epsilon\zeta a^{2} +4\epsilon\zeta$

$\frac{-1\alpha^{2}\delta^{2}}{-1\fcircle}\frac{+4\alpha^{2}a^{6}}{+4\fcircle\iota}\underline{-6\alpha^{2}a^{4}}\underline{+4\alpha^{2}a^{2}}\frac{-1\alpha^{2}}{NmedLC[隅]}$

in the unknown $A$ squared.

We can obtain the $Yo$-diagram and In-diagram for the second former and latter

equations by referring to “Case of two quartic equations” in the volume 1. But the

sentence in the process becomes _{complicated. So, for the beginners we choose the}

followingprocess. It is similar to the waythat weobtained the $Yo$-ritsu in the volume

1. You should understand that there

are

various ways to solve the problem.

The _{highest degree term in the second former equation is} $-1\alpha$ The highest

degree term in the second latter equation is $-1\alpha^{2}$ Thus, we multiply the second

former equation by $\alpha;-1\alpha\beta$ $+2\alpha\gamma$ $-1\alpha^{2}$ and subtract the product from the

second latter equation _{[by multiplying the square of} $A$ squared] to cancel the highest

degree terms. [Then,

we

get]

$+1\alpha\beta$

$-4\epsilon^{2}$ $-4\delta\zeta^{2}$ $+2\alpha^{2}\delta$

the remainder equation

$+4\alpha\epsilon a^{2} -2\alpha\gamma$ $-8\epsilon\zeta a^{2} +4\epsilon\zeta$

$\underline{-1\kappa}\underline{+4\iota}\frac{-6\alpha^{2}a^{4}}{-1\copyright}\frac{+4\alpha^{2}a^{2}}{+2\copyright}$

in the

unknown

$A$ squared.

$\overline{42This}$

sumequals that of thetermswith–sign in the first In-diagram, and theformulaat the 5th

We obtain the [converted] $Yo$

-diagram43

for the

remainder

and the second former

equations by referring “Case of two cubic equation” in the volume 1, and we

can

find

the

formulas44

of $\lambda,$

$\mu$ and $\nu$

.

By applying Cubic In-ritsu,

we

obtain the formula

Lefl

at the left-hand side and the formula which cancels out the

formula

Lefl.45

The end ofSanpo-Hakki Volume 3.

4

Sanpo-Hakki

Postscript

Sanpo-Hakkiis a book writtenbymy student Mr. IZEKI Tomotoki. Inrecent years the manuscript has been concealed in a box

as

a jewelry. $I$ looked this and laughed

the tasteless tongue. Then I made to cut printing blocks. Ah, the people who aspire

this technique get the boats and bridges to go

up,

high and far. Now the printing is

finished and I signed:

SHIMADA

Naomasa with two

seals

(

指雲，島田尚政

).

On a good morning of June $1690^{46},$

published47 by Nagano-hikouemon, $4ken$-cho, Kouraibashi-suji, Osaka

[大坂高麗橋筋四軒町 長野彦右衛門 開板].

Appendix:

Translators’ comment

on

the procedure

to

solve the

problems in the volumes 2

and

3

Seven problems

are

solved in the volumes 2 and 3. The first six problems

can

be

solved by making two equations in

an

auxiliary unknown whose coefficients contain

the true unknown. Applying the method given in the volume 1 we get

a

polynomial

equation in the true unknown. The problem

was

thought solved by this, because

they knew the celestial element method [天元術 $tengen-jutsu|$ which gives $a$ (possibly

approximate) solution of the polynomial equation with numerical coefficients if it has

a real-valued solution.

$\overline{43The}$

three equations in theunknown $y=A^{2}$ in the simplified Yo-diagramareasfollows: $44They\{$

are given in thevolume 2.

$-\lambda +\mu y -2\eta\beta y^{2} = 0$ $\mu -2\nu y +4\eta\gamma y^{2} = 0$

$-\beta +2\gamma y - \alpha y^{2} = 0$

45Thedesired answerequation in the true unknown$x$ is given in thevolume 2.

46On agood day ofMay1710inthe 2nd version.

47By Ikedaya-saburouemon, Gohuku-cho-kado, Shinsaibashi-suji, Osaka and by

Naganoya-hikouemon, $4ken$-cho, Kouraibashi-suji, Osaka in the 2nd version. One more difference is that the part)edited by IZEKI Jubeejo Tomotoki, a student of SHIMADA Naomasa” at the top of each

We get the followingtable which shows the degrees of two equations in

an

auxiliary

unknown and the degree of the final

answer

equation in the true unknown for each

problem.

The last problem No. 7 is to find the diameter from the length offive edges of a

5-lateralon acircle. Ifwehave a triangle with $B,$$d,$$e$ edgesonacircle with the diameter

$x$,

we

see that the following equation holds:

$x^{2}(2d^{2}e^{2}-d^{4}-e^{4})+\{2x^{2}(d^{2}+e^{2})-4d^{2}e^{2}\}B^{2}-x^{2}B^{4}=0.$

This is an equation in $B^{2}$ ofdegree 2 and is the first former

equation. Another same

type of equation in $B^{2}$ of degree 2 with another unknown $A^{2}$ in the coefficients is the

first latter equation. We get an equation in $A^{2}$ ofdegree 4 as a resultant. Combining

this equation with

an

equation in $A^{2}$ of degree 2 with respect to another triangle, we

get afinalequation in$x^{2}$of degree 7. So, anyexercisefor the

case

of

quintic49

equations

did not given.

Moreover, whenthedegrees of two equationsaredifferent, itis not difficult to modify

theequation_{with the higher degree toanequationof}

_one

_{degree lower without changing}

the

common

solution

as

the author used for Problem 7 in the volume 3. So, the

problems given_{in the volumes 2 and 3canbesolved byusingat highestthecaseof cubic} equations in the volume 1. In this

sense even

the real exercise for the

case

ofquartic equations did not given. This would give modem readers a little disappointment.

48The author wrote that the final equation is ofdegree 9 but theterm ofdegree 9 is actuallyzero

byourcalculation. He commented alsothatthe final equation_{in another unknown has}degree 8.